We have

$${S_k} = {{\left( {{{k - 1} \over {k!}}} \right)} \over {\left( {1 - {1 \over k}} \right)}} = {1 \over {(k - 1)!}}$$

Now, $$\sum\limits_{k = 2}^{100} {\left| {({k^2} - 3k + 1){1 \over {(k - 1)!}}} \right|} $$

$$ = \sum\limits_{k = 2}^{100} {\left| {{{{{(k - 1)}^2} - k} \over {(k - 1)!}}} \right|} $$

$$ = \sum {\left| {{{k - 1} \over {(k - 2)!}} - {k \over {(k - 1)!}}} \right|} $$

$$ = \left| {{2 \over {1!}} - {3 \over {2!}}} \right| + \left| {{3 \over {2!}} - {4 \over {3!}}} \right| + ....$$

$$ = {2 \over {1!}} - {1 \over {0!}} + {2 \over {1!}} - {3 \over {2!}} - {4 \over {3!}} + .... + {{99} \over {98!}} - {{100} \over {99!}}$$

$$ = 3 - {{100} \over {99!}}$$

Thus, $${{{{100}^2}} \over {\left| \!{\underline {\, {100} \,}} \right. }} + \sum\limits_{k = 1}^{100} {\left| {({k^2} - 3k + 1){S_k}} \right| = 3} $$